mugenziwzn6

2023-02-21

The permutations and combinations of $abcd$ taken $3$ at a time are respectively.

Helen House

Step 1. Calculate the permutations.
When n objects are taken into consideration, the number of permutations is given by,
$P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$
We have to find permutations of $abcd$ taken $3$ at a time.
$\therefore P\left(4,3\right)=\frac{4!}{\left(4-3\right)!}⇒=\frac{4×3×2×1}{1}⇒=24$
Step 2. Figure out the combination..
The number of combinations of $n$ objects taken $r$ at a time is given by,
$C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$
We have to find combinations of $abcd$ taken $3$ at a time.
$\therefore C\left(4,3\right)=\frac{4!}{3!\left(4-3\right)!}=\frac{4×3×2×1}{3×2×1×1}=4$
Hence, the permutation and combination are $24$ and $4$respectively.

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